Question

Final naïve case: If the highest-pitch string on the piano is made of spring steel (density = 7800 kg/m3) with a diameter of 1/32" (= 0.794 mm), what will the linear density of such a string be (in kg/m)?

Answer 1

Answer:

The linear density is  $K = 3.863 *10^{-3 } \ kg/m$

Explanation:

From the question we are told that

     The density of  steel is  $\rho = 7800 \ kg/m^3$

      The diameter of the string is  $d = 0.794 \ mm = 7.94 *10^{-4} \ m$

       The  radius of  the string is  evaluated as  $r = \frac{D}{2} = \frac{7.94 *10^{-4}}{2} = 3.97*10^{-4} \ m$

The volume of the string is  mathematically evaluated as

       $V = \pi * r ^2 * L$

Now assuming that the length of the string is  L = 2 m  

      So  

         $V = 3.142 * (3.97 *10^{-4})^2 * (2)$

        $V = 9.9041 *10^{-7} \ m^3$

Then the mass of the string would be  

       $m = \rho * V$

substituting value  

       $m = 7800*9.904 14 *10^{-7}$

      $m = 7.73*10^{-3} \ kg$

Looking at the question we see that the unit of the linear density is  $\frac{kg}{m}$

Hence the linear density is evaluated as

        $K = \frac{m}{L}$

substituting value  

        $K = \frac{7.73 *10^{-3}}{2}$

        $K = 3.863 *10^{-3 } \ kg/m$